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  • 10-07-2021
  • Mathematics
contestada

The numbers a, b and c satisfy a+b+c = 0 and abc = 78. What is
the value of (a+b)(b+c)(c+a)?​

Respuesta :

lublana
lublana lublana
  • 15-07-2021

Answer:

[tex](a+b)(b+c)(c+a)=-78[/tex]

Step-by-step explanation:

We are given that the numbers a, b and c

[tex]a+b+c=0[/tex]   ....(1)

[tex]abc=78[/tex]    ..... (2)

We have to find the value of (a+b)(b+c)(c+a).

From equation we get

[tex]a+b=-c[/tex]

[tex]b+c=-a[/tex]

[tex]a+c=-b[/tex]

Substitute the values then we get

[tex](a+b)(b+c)(c+a)=(-c)(-a)(-b)[/tex]

[tex](a+b)(b+c)(c+a)=-abc[/tex]

Using equation (2) we get

[tex](a+b)(b+c)(c+a)=-78[/tex]

Hence, the value of [tex](a+b)(b+c)(c+a)[/tex] is -78.

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